If f(x) is a continuous function in `[0,pi]` such that f(0)=f(x)=0, then the value of
`int_(0)^(pi//2) {f(2x)-f''(2x)}sin x cos x dx` is equal to

Options

A. `pi`
B. `2pi`
C. `3pi`
D. 0

Answer

D

Explanation

Let`I=underset(0)overset(pi//2)int {f(2x)+f''(2x)}sin xcos x dx`.Then,
`I=(1)/(2)underset(0)overset(pi//2)int {f(2x)+f''(2x)}sin 2x dx`
`rArr I=(1)/(2)underset(0)overset(pi)int{f(t)+f''(t)} sin t dt`, where t=2x
`I=(1)/(2)underset(0)overset(pi)int f underset(I)((t))underset(II)sin t dt +(1)/(2)underset(0)overset(pi)int f''underset(II)((t))underset(I)sin t dt `
`rArr I=(1)/(2)[[-f(t)cos t]_(0)^(pi)+underset(0)overset(pi)int f'(t)cos t dt+[-f'(t)sin t]_(0)^(pi)-underset(0)overset(pi)int f'(t)cos t dt]`
`I=0`